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STD 11 - 12. limits — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 12. limits4 Marks
MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}} = $
  • A
    $0$
  • B
    $1$
  • C
    $-1$
  • ✓
    Does not exist
✓

Answer

Correct option: D.
Does not exist
d
(d) $f(x) = \left( {\frac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}}} \right)\,,$ then

$\mathop {\lim }\limits_{x \to \,0 + } \,f(x) = \mathop {\lim }\limits_{h \to \,0} \left( {\frac{{{e^{1/h}} - 1}}{{{e^{1/h}} + 1}}} \right) $

$= \mathop {\lim }\limits_{h \to \,0} \frac{{{e^{1/h}}\left( {1 - \frac{1}{{{e^{1/h}}}}} \right)}}{{{e^{1/h}}\left( {1 + \frac{1}{{{e^{1/h}}}}} \right)}} = 1$

Similarly $\mathop {\lim }\limits_{x \to \,0 - } f(x) = - 1$. 

Hence limit does not exist.

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